Problem: Find one value of $x$ that is a solution to the equation: $(x^2-2)^2-10(x^2-2)+21=0$ $x=$
Explanation: We could solve for $x$ by expanding $(x^2-2)^2$ and $-10(x^2-2)$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Note that if we let ${p}={x^2-2}$, we can rewrite the equation: $({x^2-2})^2-10({x^2-2})+21=0$ In particular, we can express it in the form: ${p}^2-10{p}+21=0$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2-10{p}+21&=0\\\\ ({p}-7)({p}-3)&=0\\\\ {p}=7\ &\text{or} \ \ {p}=3 \end{aligned}$ Since ${p}={x^2-2}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${x^2-2}=7\ \ \ \text{or} \ \ \ {x^2-2}=3$ When we solve ${x^2-2}=7$, we find that $x=\pm3$. When we solve ${x^2-2}=3$, we find that $x=\pm\sqrt{5}$. In conclusion, the four solutions of the equation $(x^2-2)^2-10(x^2-2)+21=0$ are: $x=3$ $x=-3$ $x=\sqrt{5}$ $x=-\sqrt{5}$ [Is there another way to solve for x?]